The resultant of `vecA` and `vecB` is perpendicular to `vecA`. What is the angle between `vecA` and `vecB` ?

Options

A. `cos^(-1)((A)/(B))`
B. `cos^(-1)(- (A)/(B))`
C. `sin^(-1)((A)/(B))`
D. `sin^(-1)(-(A)/(B))`

Answer

2

Explanation

`tan alpha = (B sin theta )/( A+ B cos theta ) = tan 90^(@) = oo`
thus, `A+B costheta =0 rArr cos theta = - A//B`